4
$\begingroup$

I am still struggling quite a lot with the Picard-Lindelöf-Theorem (also known as the Cauchy-Lipschitz-Theorem).

Problem: Consider the following IVP with $\alpha \neq 1$ $$\begin{cases} y'&= t|y|^\alpha \\ y(0)&=1 \end{cases} $$ And show that there exists a unique solution $f_\alpha$ on an Interval $I_\alpha$ with $0 \in I_\alpha$, also show that $f_\alpha (t) > 0$ for small $t$

So to use Picard-Lindelöf I want to show that $f(t,y)=t|y|^\alpha$ is Lipschitz-continuous with respect to its second variable. I could do this by computing the partial derivative with respect to $y$ and find an interval on which this expression is bound, then on said hypothetical interval the solution would exist and be unique.


My approach: $$\frac{\partial f}{\partial y}= t\alpha|y|^{\alpha-1} \cdot \frac{y}{\sqrt{y^2}} $$ At which point I already run into trouble. I have made use of the fact that $|x|'= x/\sqrt{x^2}$ because it gives correct results in regard of the direction (when approaching from the left and the right). But the above expression is not defined at $y=0$.

So I thought if I could come up with an argument that shows why $y \neq 0$, then at least this problem would be out of the way.

Assume that $y=0$ then $y'=0=f(t,y) \implies \frac{\partial f}{\partial y}=0$ which is bound on $\mathbb{R} \implies f$ is Lipschitz continuous with respect to $y$ but we have $y(0)=1$ and thus $$y'=0 \implies y=c $$ So $y$ is a constant function and $y(0)=1$ means that $c=1 \implies y=1 \neq 0$ which shows that $y$ can not be the 'zero' function (if that exists).

Questions:

  • Is the above argumentation correct?
  • How can I find the desired interval $I_\alpha$ with $0 \in I_\alpha$?

Update: I tried to continue a bit on my own and ignore the Picard-Lindelöf part and just focus on solving the IVP, apparently the step I have taken above was necessary so I can divide by $|y|^\alpha$, however I will land at $$\frac{dy}{|y|^\alpha}=tdx $$ So it seems like I have to come up with a condition that guarantees me that $y$ is positive in some Intervall.

$\endgroup$
1
$\begingroup$

The "argumentation" can hardly be called such; it's an inconclusive train of thought. So I'll leave it aside and address the problem: what does Picard-Lindelöf theorem actually say?

It says: consider the problem $y'=f(t,y)$ with $y(t_0)=y_0$, and suppose $(t_0,y_0)$ has a neighborhood (a rectangle $R$ in $ty$-plane) in which $f$ is Lipschitz in $y$ and continuous in $t$. Then there is an interval $(t_0-\delta,t_0+\delta)$ in which a solution exists, is unique, and stays within the aforementioned rectangle.

In your case, $(t_0,y_0)=(0,1)$. Take $R$ to be $[-1,1]\times [1/2,3/2]$; all that matters is that we stay away from $y=0$. Check that $f$ satisfies the above properties on $R$; this isn't hard especially because $|y|=y$ (we have positive values of $y$ only). Apply the theorem. Done.

Nothing in the problem statement says that you have to produce an explicit interval of existence/uniqueness. But if you want to do this, it's best to use the optimized version which gives the interval of existence/uniqueness as $|t-t_0|\le \min(a,b/M)$. Here $a$ and $b$ describe the rectangle $R$ that is ours to choose; with my choice $a=1$ and $b=1/2$. The number $M$ is the maximum of $|f|$ on said rectangle, which turns out to be $(3/2)^\alpha$. So, in the interval $|t|\le 2^{-1}(2/3)^{\alpha}$ the solution exists and is unique.

$\endgroup$
-1
$\begingroup$

You should work on the specific values of \alpha. For example, if \alpha >1, then we get the local existence of non-negative solution, which is blow-up at a certain time. It is of course that you also get the uniqueness in this case. Other way, I propose you a different approach to your problem. You can approximate |y|^\alpha by (|y|+\delta)^\alpha. Then, you have no problem with the equation of y_\delta. After that you try to pass to the limit as \delta\rightarrow 0 in order to get the result.

Good luck

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.